Question

The person first reaches a height of 200 feet after traveling a horizontal distance between
feet.

Answer 1

I don't have enough information to get a numeric answer — the problem statement is missing the equation or the launch data. Could you paste the full problem or give the height function or initial speed/angle?

Meanwhile, here is how to solve it once you have the trajectory:

1. If the height as a function of horizontal distance x is y(x), set y(x) = 200.
2. Solve that equation for x. If it is quadratic (typical for projectile motion), you'll get two roots; the smaller positive root is the horizontal distance where the person first reaches 200 ft.
- For a quadratic ax^2 + bx + c = 0 with c adjusted so the equation is ax^2 + bx + (c−200)=0, use the quadratic formula x = [−b ± sqrt(b^2 − 4a(c−200))]/(2a).
3. Report the smaller positive root (if any). If there is no real positive root, the person never reaches 200 ft.

Example (illustration): if y(x) = −0.02x^2 + 4x + 10, set −0.02x^2 + 4x + 10 = 200 → −0.02x^2 + 4x −190 = 0. Multiply by −50: x^2 −200x + 9500 = 0. Then x = [200 ± sqrt(200^2 − 4·9500)]/2 = [200 ± sqrt(40000 − 38000)]/2 = [200 ± sqrt(2000)]/2 ≈ [200 ± 44.72]/2. The first (smaller) positive root ≈ (200 − 44.72)/2 ≈ 77.64 ft.

If you post the specific function or the rest of the problem I’ll compute the exact numeric answer.